I would like to import a function from the views.py file of my import module.
The following code does not work: from import.views import myFunction. How to do?
**
Edit:
** So far I was able to import other files from the import module:
sys.path.append(settings.SITE_ROOT+'import')
import myFile as test
It doesn't seem to work for the view though, don't know why!
Related
I was trying to create some steps in the Abaqus by using the following python code. Unfortunately having this error. Anybody, please help me...
KeyError:model_name
Python Code:
from abaqus import *
from abaqusConstants import *
import __main__
import section
import regionToolset
import displayGroupMdbToolset as dgm
import part
import material
import assembly
import step
import interaction
import load
import mesh
import optimization
import job
import sketch
import visualization
import xyPlot
import displayGroupOdbToolset as dgo
import connectorBehavior
def create_step(model_name, new_step, previous_step):
mdb.models['model_name'].StaticStep(name='new_step', previous='previous_step', initialInc=0.025,
maxInc=0.025)
session.viewports['Viewport: 1'].assemblyDisplay.setValues(step='new_step')
model_name = 'Model-' + str(0)
new_step = 'C4'
previous_step = 'C3'
create_step(model_name, new_step, previous_step)
Replace mdb.models['model_name'].Stat... with mdb.models[model_name].Stat...
def create_step(model_name, new_step, previous_step):
mdb.models['model_name'].StaticStep(name='new_step', previous='previous_step', initialInc=0.025,
maxInc=0.025)
session.viewports['Viewport: 1'].assemblyDisplay.setValues(step='new_step')
2nd line should be,
mdb.models[model_name].StaticStep(name='new_step', previous='previous_step', initialInc=0.025,
maxInc=0.025)
I don't know what happened but all of the sudden i am getting this error:
Invalid template library specified. ImportError raised when trying to load 'bootstrap3.templatetags.bootstrap3': cannot import name 'flatatt'
any ideas?
Also you can get the latest version of django-bootstrap3 . That fixed it for me.
look for the file which gives the error
and change the line having flatatt in the import to the given line
from django.forms.utils import flatatt
I had also such problem and I found the way like this!
1-step) in components.py file in venv\Lib\site-packages\bootstrap3
it was written :
from django.forms.widgets import flatatt
you should change it to
from django.forms.utils import flatatt
2-step)in utils.py file in folder venv\Lib\site-packages\bootstrap3
it was written :
from django.forms.widgets import flatatt
you should change it to :
from django.forms.utils import flatatt
I compiled my prototype Application with py2exe to check its function as an exe, and run into 0 errors until I go to start it. Nothing happens. A process starts with my app name, it thinks for a few seconds, then nothing. No log file is generated. The app works great when run in python environment, but not in the compiled exe. I've given my setup code below. Any ideas? :
from distutils.core import setup
import py2exe, sys, os
import matplotlib
import FileDialog
import dateutil
sys.argv.append('py2exe')
setup( windows=['ATLAS.pyw'], data_files=matplotlib.get_py2exe_datafiles(),
options = {"py2exe": {
"includes": "decimal, datetime",
"packages": ["FileDialog", "dateutil"],
'bundle_files': 2,
'compressed': True}
},
zipfile = None
)
Hooks utilized in the Application:
from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg, NavigationToolbar2TkAgg
from matplotlib.backend_bases import key_press_handler
from pandas.sandbox.qtpandas import DataFrameWidget
from matplotlib.widgets import LassoSelector
from tkFileDialog import askopenfilename
from matplotlib.figure import Figure
import matplotlib.image as mpimg
from PySide import QtGui, QtCore
from matplotlib.path import Path
import pandas.io.sql as psql
from numpy import nonzero
import tkMessageBox as mb
from pylab import *
import pyodbc
import sys
import ttk
SOLVED:
So, using quick fingers (and compiling it with PyInstaller with the --debug option) I screen-capped the quickly-closing console window that contained the Traceback:
WindowsError: [Error 3] The system cannot find the path specified: 'C:\\path\\dateutil\\zoneinfo/*.*'
The zoneinfo file was being saved in pytz instead of dateutil. A quick rename solved the problem.
Only issue though, if you want to compile with -F or --onefile it will not work due to the initial improper naming convention. Not quite sure how to fix that though.
I have hundreds of routes in my flask main module,
I think it need to separate those hundred of routes from the main module.
How to do it ?
#!/usr/bin/env python3
# -*- coding: utf8 -*-
from flask import request, url_for
from flask import Flask, request, jsonify
from flask_request_params import bind_request_params
from flask import g
import datetime
import pandas as pd
import pymongo
from webargs import Arg
from webargs.flaskparser import use_args, use_kwargs
import yaml
import time, functools
from pdb import set_trace
from pandas_helper import PandasHelper
import errors
from app_helper import *
from release_schedule import ReleaseSchedule
from mongo import Mongo
#app.route('/next_release', methods=["GET"])
#return_json
def next_release():
schedules = ReleaseSchedule.next_release(DB)
return pd.DataFrame([sche for sche in schedules])
...
#app.route('/last_release', methods=["GET"])
This is what blueprints were made to do.
Another alternative is flask-classy (which is awesome). I'm going to talk about the blueprint approach since that's what I know better.
If I was in your position I would want to split my routes up based on common imports.
Without knowning your application I'm going to guess that a distribution like this
parse_user_data_views.py
from webargs import Arg
from webargs.flaskparser import use_args, use_kwargs
import yaml
push_to_db_views.py
from pandas_helper import PandasHelper
from mongo import Mongo
import pymongo
import pandas as pd
import datetime
release_views.py
from release_schedule import ReleaseSchedule
import pandas as pd
#app.route('/next_release', methods=["GET"])
#return_json
def next_release():
schedules = ReleaseSchedule.next_release(DB)
return pd.DataFrame([sche for sche in schedules])
is likely distribution. We can't answer this for you, only you can.
But this allows you to separate out your application in some pretty nice ways.
in __init__.py
from flask import Flask
from yourapplication.release_views import release_views
from yourapplication.push_to_db_views import push_to_db_views
from yourapplication.parse_user_data_views import parse_user_data_views
app = Flask(__name__)
app.register_blueprint(release_views)
app.register_blueprint(push_to_db_views)
app.register_blueprint(parse_user_data_views)
Create a new file called views.py and add all your routes there. Then import views.py in your __ init __.py .
I am working on a Django project.
I currently import functions, models like:
from project.abc.models import *
from project.cba.models import *
from project.abc.views import *
from project.cba.views import *
Is there any syntax which can enable me to write just the name of project and it may return every model, functions and etc from every application like:
from project.models import *
from project.views import *
from project.urls import *
I am using Django1.3, and I know that it is not a good practice to import * of anything, but this is my need at the moment. Please help!
I think that it is not a good idea but if you want you can create a file models.py in your project root directory and write there:
from project.abc.models import *
from project.cba.models import *
after that you can import your models like:
from project.models import SomeModel
And you also can create views.py etc.